To solve the limit

$\lim_{t \to 0} \left( \frac{1}{t \sqrt{1 + t}} - \frac{1}{t} \right),$

we'll first try to simplify the expression inside the limit.

Step 1: Combine the Fractions

Rewrite the expression inside the parentheses by finding a common denominator:

$\frac{1}{t \sqrt{1 + t}} - \frac{1}{t} = \frac{1 - \sqrt{1 + t}}{t \sqrt{1 + t}}.$

Step 2: Simplify the Numerator

Now, let's focus on simplifying $1 - \sqrt{1 + t}$. To do this, we can multiply the numerator and the denominator by the conjugate, $1 + \sqrt{1 + t}$:

$1 - \sqrt{1 + t} = \frac{(1 - \sqrt{1 + t})(1 + \sqrt{1 + t})}{1 + \sqrt{1 + t}} = \frac{1 - (1 + t)}{1 + \sqrt{1 + t}} = \frac{-t}{1 + \sqrt{1 + t}}.$

Thus, we can rewrite the expression as

$\frac{1 - \sqrt{1 + t}}{t \sqrt{1 + t}} = \frac{-t}{t \sqrt{1 + t} (1 + \sqrt{1 + t})}.$

Step 3: Cancel Terms

Now, cancel $t$ in the numerator and denominator:

$\frac{-t}{t \sqrt{1 + t} (1 + \sqrt{1 + t})} = \frac{-1}{\sqrt{1 + t} (1 + \sqrt{1 + t})}.$

Step 4: Take the Limit as $t \to 0$

Now we can evaluate the limit by substituting $t = 0$:

$\lim_{t \to 0} \frac{-1}{\sqrt{1 + t} (1 + \sqrt{1 + t})} = \frac{-1}{\sqrt{1} (1 + \sqrt{1})} = \frac{-1}{1 \cdot 2} = -\frac{1}{2}.$

Final Answer

$\lim_{t \to 0} \left( \frac{1}{t \sqrt{1 + t}} - \frac{1}{t} \right) = -\frac{1}{2}.$

Would you like further details, or have any questions?

1. What is the purpose of multiplying by the conjugate when simplifying expressions?
2. How do we know when terms can be canceled in fractions?
3. What are other methods to approach limits with indeterminate forms?
4. How would this problem change if it were $\lim_{t \to 0^+}$?
5. Why do we evaluate $\sqrt{1 + t}$ as $t \to 0$?

Tip: Multiplying by the conjugate is a powerful technique for handling limits involving square roots and allows you to eliminate square roots in the numerator.